´AGOTA OROSZ Received 10 July 2005; Revised 24 October 2005; Accepted 30 October 2005 The classical theory of homogeneous and inhomogeneous linear difference equations with constant coeffic
Trang 1´AGOTA OROSZ
Received 10 July 2005; Revised 24 October 2005; Accepted 30 October 2005
The classical theory of homogeneous and inhomogeneous linear difference equations with constant coefficients on the set of integers or nonnegative integers provides effective solution methods for a wide class of problems arising from different fields of applications However, linear difference equations with nonconstant coefficients present another im-portant class of difference equations with much less highly developed methods and theo-ries In this work we present a new approach to this theory via polynomial hypergroups
It turns out that a major part of the classical theory can be converted into hypergroup language and technique, providing effective solution methods for a wide class of linear difference equations with nonconstant coefficients
Copyright © 2006 ´Agota Orosz This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
1 Introduction
A linear difference equation with nonconstant coefficients has the following general form:
aN(n) fn+N+aN −1(n) fn+N −1+···+a1(n) fn+1+a0(n) fn = gn, (1.1) where the functionsa0,a1, , aN,g : N → Care given withaN not identically zero, and
N, k are fixed nonnegative integers (in this paper N = {0, 1, 2, }) The above equation is supposed to hold for some unknown function f : N → Cor f : Z → C, depending on the nature of the problem In what follows we will prefer the case f : N → Cand the notation
f (m) and g(m) instead of fmandgm
By the classical theory of differential equations the solution space of the above
equa-tion can be described completely in the constant coe fficient case, that is, if the functions
a0,a1, , aN are constants In this case the solution space is generated by exponential monomial solutions, which arise from the roots of the characteristic polynomial, called characteristic roots.
Hindawi Publishing Corporation
Advances in Di fference Equations
Volume 2006, Article ID 51427, Pages 1 10
DOI 10.1155/ADE/2006/51427
Trang 2Much less is known in the case of nonconstant coefficients In this work we offer a method to solve some types of homogeneous linear difference equations with noncon-stant coefficients by transforming these equations into homogeneous linear difference equations with constant coefficients over hypergroups This method is based on some
theory of homogeneous linear difference equations with constant coefficients on hyper-groups developed along the lines of the classical theory overN The basic idea is that the role of exponential functions is played by the generating polynomials of some polynomial hypergroups
Some results of this work have been presented at the 5th Debrecen-Katowice Winter Seminar in Be¸dlewo (Poland) in 2005
We remark that most of this work can be generalized to the case of signed hypergroups,
as they are presented in [1] Nevertheless, in the forthcoming presentation we restrict ourselves to polynomial hypergroups in the sense of [2]
2 Discrete polynomial hypergroups
Let (α n)n ∈N, ( β n)n ∈Nand (γ n)n ∈Nbe real sequences with the following properties:γ n > 0,
β n ≥0,α n+1 > 0 for all n inN, moreoverα0=0, andα n+β n+γ n =1 for alln inN We define the sequence of polynomials (Pn)n ∈NbyP0(x) =1,P1(x) = x, and by the recursive
formula
xPn(x) = αnPn −1(x) + βnPn(x) + γnPn+1(x) (2.1) for alln ≥1 andx inR In this case there exists constantsc(n, m, k) for all n, m, k inN
such that
P n P m =
n+m
k =| n − m |
holds for alln, m inN(see [3,4]) This formula is called linearization formula, and the
coefficients c(n,m,k) are called linearization coefficients It is clear that P n(1)=1 for alln
inN, hence we have
n+m
k =| n − m |
for alln inN If the linearization coefficients are nonnegative: c(n,m,k)≥0 for alln, m, k
inN, then we can define a hypergroup structure onNby identifying the natural numbers with the Dirac-measures in virtue of the following rule:
δ n ∗ δ m =
n+m
k =| n − m |
for alln, m inN, with involution as the identity mapping and withe as 0 The resulting
hypergroupK =(N,∗ ) is called the polynomial hypergroup associated with the sequence
(Pn)n ∈N
Trang 3Let f : N → Cis an arbitrary function andm is a natural number The translate of f by
m is defined by
᐀m f (n) =
Nf (k)d
δn ∗ δm
for alln inN Althoughn ∗ m is not defined on the hypergroup, the following notation
is in use:
f (n ∗ m) =᐀m f (n) =
n+m
k =| n − m |
for eachn, m inN The functionχ : N → C is said to be an exponential function on the
polynomial hypergroup if
holds for alln, m inN If the hypergroup is generated by the sequence of polynomials (Pn)n ∈N, then a functionχ : N → Cis an exponential function if and only if
holds for some complex numberλ (see [2])
3 Di fference equations with 1-translation
In the classical theory of difference equations the translate of a function by n and the
translation of the functionn-times by 1 give the same result for all n inN But in the hypergroup case there are two different ways to define difference equations along these two interpretations In this section we deal with the latter one We introduce the notation
for anyf : N → Candn inN, moreover᐀0f = f and᐀N f =᐀(᐀N −1 f ) for each integer
N > 1 Obviously,᐀ is a linear operator on the linear spaceC Nof all complex valued functions onN IfQ is any polynomial with complex coe fficients, then Q(᐀) has the
obvious meaning LetN be a positive integer, a0, , aNbe complex numbers and suppose thata N =0 We will consider functional equations of the form
Q( ᐀) f = a N᐀N f (n) + a N −1᐀ N −1 f (n) + ···+a0f (n) =0, (3.2)
which is called a homogeneous linear di fference equation of order N on the hypergroup K with constant coefficients associated to the polynomial Q The polynomial Q is called the characteristic polynomial of (3.2) and its roots are called the characteristic roots of (3.2) The solution space of (3.2) is the kernel of the linear operatorQ(᐀), hence it is a linear subspace of the function spaceC N This solution space is translation invariant in the sense that if f is a solution, then ᐀ f is a solution, too.
Trang 4Theorem 3.1 If Q is a complex polynomial of degree N ≥ 1, then the solution space of ( 3.2) has dimension N.
Proof Suppose that f : N → Cis a solution of (3.2) Since f (0 ∗1)= f (1) and for n ≥1
we have
f (n ∗1)=
n+1
k = n −1
c(n, 1, k) f (k) = α n f (n −1) +β n f (n) + γ n f (n + 1), (3.3)
where (α n)n ∈N, (β n)n ∈Nand (γ n)n ∈Nare the sequences appearing in the definition of the polynomial hypergroup Considering (3.2) forn =0 we get
aN γN −1··· γ1γ0f (N) +
N−1
i =0
with some complex numbersk N,i (i =0, , N −1) Since obviouslya N γ N −1 ··· γ0=0, hence f (N) is determined by f (0), , f (N −1) and it is easy to see by induction that
f (n) is uniquely determined by the values f (0), , f (N −1) forn ≥ N.
Theorem 3.2 If the complex number λ is a characteristic root of (3.2) with multiplicity m, then all the functions n → P(n k)(λ) are solutions of (3.2) for k =0, 1, , m − 1.
Proof By the exponential property of the function n → P n(λ) we have
᐀P n(λ) = P n(λ)P1(λ) = λP n(λ), ᐀t P n(λ) = λ t P n(λ), (3.5) thus we can see immediately that this function is a solution of (3.2):
Q( ᐀)P n(λ) =λ N+aN −1λ N −1+···+a1λ + a0
Pn(λ) =0. (3.6) For proving thatn → P n(k)(λ) are also solutions for 1 ≤ k ≤ m, we need the translates of
P(n k)(λ) After some calculation we get that
᐀r P(k)
n (λ) =
min( r,k)
t =0
r t
λ r − t k!
(k − t)! P
(k − t)
forr inN, therefore we have
Q( ᐀)P(k)
n (λ) =
N
r =0
a r᐀r P(k)
n (λ)
=
N
r =0 ar
min(r,k)
t =0
r t
λ r − t k!
(k − t)! P
(k − t)
n (λ)
=
k
t =0
k t
N
r = t
ar r!
( − t)! λ
r − t
P n(k − t)(λ) =0,
(3.8)
as the tth derivative of the characteristic polynomial at λ is equal to zero for 0 ≤ t ≤
Trang 5Lemma 3.3 Let k be a positive integer and l1, , lk nonnegative integers If λ1, , λk are different complex numbers, then the functions P(i)
n (λ j ) are linearly independent for j =
1, 2, , k and i =0, 1, , lj
Proof First we show that Pn(λ1), , Pn(λk) are linearly independent ifλ1, , λkare di ffer-ent If it is not the case, then there are complex numbersa1,a2, , ak, not all equal to zero, with the propertyk
i =1 aiPn(λi)=0, which contradicts the fact that for some constantC
the following equation holds:
P0
λ1
··· P0
λk
P1
λ1
··· P1
λk
P k
λ1
··· P k
λ k
= C
λ1 ··· λ k
λ k1 ··· λ k k
Now assume that there existλ1, , λkdifferent complex numbers such that the func-tionsn → P(n i)(λ j) are linearly dependent forj =1, 2, , k and i =0, 1, , l jfor some pos-itive integersl1, , lk Suppose thatk is the minimal positive integer with this property,
and also suppose thatl1+···+l kis minimal It means, that there exist complex numbers
aj,i, not all equal to zero forj =1, , k and i =0, , l jsuch that
k
j =1
l j
i =0
a j,i P(i) n
λ j
holds withaj,l j =0 Translating (3.10) by 1 we have
k
j =1
l j
i =1 aj,i
λ jP(n i)
λj
+P(n i −1)
λ j
+
k
j =1 aj,0λj Pn
λ j
and if we subtract (3.10) timesλ1from this equation we get an expression which does not containP(l1 )
n (λ1):
l 1−1
i =0
c1,i P(i)
n
λ1
+
k
j =2
a j,l j
λ1− λ j
P(l j)
n
λ j
+
k
j =2
lj −1
i =0
c j,i P(i) n
λ j
with some constantscj,i, and this means that eitherk or l1+···+lk was not minimal
Using Theorems3.1,3.2andLemma 3.3we can characterize the solution space of (3.2) completely
Theorem 3.4 Let Q be a complex polynomial of degree N ≥ 1 with all di fferent complex zeros λ1,λ2, , λk, where the multiplicity of λ j is lj (j =1, 2, , k) Then the function f :
N → C is a solution of (3.2) if and only if it is a linear combination of functions of the form
n → P(n i)(λj ) with j =1, 2, , k and i =0, 1, , lj − 1.
Trang 64 Difference equations with general translation
Let us consider the following equation for a function f : N → C
a N᐀N f (n) + a N −1᐀ N −1 f (n) + ···+a0f (n) =0, (4.1) whereN is a positive integer and aN, , a0are complex numbers We note, that (4.1) can
be written in the form
aN f (n ∗ N) + aN −1f
n ∗(N −1)
+···+a0f (n) =0. (4.2)
It is easy to see, that the solution space of (4.1) is a linear subspace ofC Nwith dimension
N We will show, that this solution space is generated by similar functions like in the case
of (3.2), but the characteristic polynomial is different: it depends on the basic generating polynomials of the hypergroup
Theorem 4.1 The function f : N → C is a solution of (4.1) if and only if it is the linear combination of functions of the form n → P n(i)(λ j ) with j =1, 2, , k and i =0, 1, , l j − 1, where λ1,λ2, , λ k are di fferent complex zeros of the polynomial
λ −→ aNPN(λ) + aN −1PN −1(λ) + ···+a1P1(λ) + a0, (4.3)
and the multiplicity of λ j is lj(j =1, 2, , k).
Proof It will be su fficient to show, that the functions n → P n(i)(λ j) with j =1, 2, , k and
i =0, 1, , l j −1 are solutions Since
᐀m
P(n i)(λ)
=
i
t =0
i t
P(t)(λ)P n(i − t)(λ) (4.4)
for allm inN, substitutingP(n i)(λ) instead of f (n) in (4.1) we get
N
m =0
am᐀m
P(i)
n (λ)
=
N
m =0 am i
t =0
i t
P(t)(λ)P(i − t)
n (λ)
=
i
t =0
i t
P(i − t)
n (λ)
N
m =0
a m P(t)(λ)
=0,
(4.5)
which holds ifλ is a root of (4.3) with a multiplicity higher thani.
5 Examples
Example 5.1 We consider the equation
On the Chebyshev-hypergroup we have
᐀ f (n) =1
2
f (n + 1) + f
Trang 7hence (5.1) has the form
f (n + 1) + f
forn =0, 1, With n =0 we have f (1) =0, and withn ≥0 it follows f (n + 2) + f (n) =
0, which impliesf (2n + 1) =0 and f (2n) =(−1)n f (0) On the Legendre-hypergroup we
have
᐀ f (n) = n + 1
2n + 1 f (n + 1) +
n
2n + 1 f
hence (5.1) has the form
(n + 1) f (n + 1) + n f
forn ≥0 Withn =0 we have f (1) =0, and withn ≥0 it follows (n + 2) f (n + 2) + (n +
1)f (n) =0, which impliesf (2n + 1) =0, moreoverf (2n) =(−1)n((2n −1)!!/(2n)!!) f (0).
(Heren!! denotes the double factorial of n.)
One observes that in the first case f (n) = f (0) · T n(0) and in the second case f (n) =
f (0) · Pn(0), whereTn, respectivelyPn denotes thenth Chebyshev-polynomial,
respec-tively thenth Legendre-polynomial This is a simple consequence of our previous results.
Indeed, the characteristic polynomial of (5.1) isQ(λ) = λ, hence the only characteristic
root isλ =0 with multiplicity 1 Hence, on any polynomial hypergroup with generat-ing polynomials (Pn)n ∈NbyTheorem 3.4, the general solution of the difference equation (5.1) has the form f (n) = f (0) · P n(0)
Now we consider the following problem: find all solutions f : N → Cof the difference equation
(n + 2) f (n + 2) −(2n + 3) f (n + 1) + (n + 1) f (n) =0 (5.6) with f (0) = f (1) Observe, that by introducing g(n) =(n + 1) f (n + 1) −(n + 1) f (n) for
n =0, 1, we have g(n + 1) − g(n) =0, which means thatg is constant and g(n) = g(0) =
f (1) − f (0) =0 It follows (n + 1) f (n + 1) −(n + 1) f (n) =0, which implies again that f
is constant: f (n) = f (0) for each n inN Then again one can realize that (5.6) is exactly the difference equation
on the Legendre-hypergroup, which is a very special case of (3.2), and can be solved with the method we offered above Indeed, the characteristic polynomial has the form
Q(λ) = λ −1 and the only characteristic root isλ =1 with multiplicity 1 According to
Theorem 3.4, the general solution of the difference equation (5.6) has the form f (n) =
f (0) · Pn(1), wherePnis thenth Legendre-polynomial As Pn(1)=1 for eachn inN, we have that all solutions of (5.6) satisfyingf (0) = f (1) are constant.
We can modify (5.7) to consider
Trang 8wherec is a complex parameter This is the eigenvalue problem for the translation
oper-ator᐀ on any polynomial hypergroup with generating polynomials (P n)n ∈N In this case the characteristic polynomial isQ(λ) = λ − c having the only characteristic root λ = c
with multiplicity 1 Hence each complex numberc is an eigenvalue with the
correspond-ing eigenfunctionn → Pn(c).
Example 5.2 The study of higher order difference equations leads naturally to the study
of generalized polynomial functions on polynomial hypergroups We will consider this problem in more details elsewhere, here we work out a simple special case only Consider the difference equation
orΔ2f =0, where we use the notation Δ=᐀− I, and I is the identity operator The
generating polynomials of the underlying polynomial hypergroup are the polynomials (P n)n ∈N The characteristic polynomial of (5.9) isQ(λ) = λ2−2λ + 1 =(λ −1)2, hence the only characteristic root isλ =1 with multiplicity 2 ByTheorem 3.4the general solution
of (5.9) has the form
f (n) = A · P n(1) +B · P n(1)= A + B · P n(1), (5.10) with arbitrary complex constants A, B We know from the results of [5] that n → B ·
P n(1) represents a general additive function on the given hypergroup, hence (5.9) can be considered as a characterization of affine functions on polynomial hypergroups—exactly
as in the group case
Example 5.3 Our last example illustrates the application ofTheorem 4.1 We consider the difference equation
on the hypergroup which is generated by the sequences (αn)n ∈N, (βn)n ∈Nand (γn)n ∈N By
Theorem 4.1the general solution of (5.11) can be described with the help of the roots of the polynomial
whereP ndenotes thenth basic polynomial for all n inN Using the recursive formula for
n =1 and the propertyαn+βn+γn =1 we get
γ1
P2(λ) −2P1(λ) + 1
=(λ −1)
λ −γ1− α1
hence the solutions are the functions
f (n) = APn(1) +BPn
γ1− α1
= A + BPn
γ1− α1
(5.14)
Trang 9with arbitrary complex numbersA, B On the Chebyshev-hypergroup, where the
recur-sive formula for the Chebyshev-polynomials (Tn)n ∈Nis
λTn(λ) =1
2Tn+1(λ) +
1
withT0(λ) =1 andT1(λ) = λ the above equation has the form
f (n + 4) −2f (n + 3) + 2 f (n + 2) −2f (n + 1) + f (n) =0 (5.16) with the initial conditions
f (2) =2f (1) − f (0), f (3) = f (1). (5.17) The general solution has the form
with arbitrary complex numbersA, B, and more explicitly, we can write f (2n) = A + B( −1)nand f (2n + 1) =0 for eachn inN As in this case the problem reduces to a linear homogeneous difference equation with constant coefficients, the same result can be de-rived from the classical theory Nevertheless, in the case of the Legendre-hypergroup one obtains a linear homogeneous difference equation with nonconstant coefficients, and the classical methods cannot be directly applied but by the virtue of (5.14) we know that the solutions are the functions
f (n) = A + B · Pn 1
3
where P n is the nth Legendre-polynomial Similarly, in the case of the
Chebyshev-hypergroup of the second kind the recursive formula has the form
λU n(λ) = n + 2
2n + 2 U n+1(λ) +
n
2n + 2 U | n −1|( λ) (5.20)
withU0(λ) =1 andU1(λ) = λ, thus the solutions of (5.11) are the functions
f (n) = A + B · Un 1
2
Acknowledgment
The research was supported by the Hungarian National Foundation for Scientific Re-search (OTKA), Grant no T-016846
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Trang 10[2] W R Bloom and H Heyer, Harmonic Analysis of Probability Measures on Hypergroups, de
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´Agota Orosz: Institute of Mathematics and Informatics, University of Debrecen, 4010 Debrecen, Hungary
E-mail address:oagota@math.klte.hu